65

Let me quote von Neumann twice: There's no sense in being precise when you don't even know what you're talking about.   Young man, in mathematics you don't understand things. You just get used to them. The notion of measure extends the notion of length for intervals, or area for "simple enough shapes" in the plane, or volume of "simple enough ...


36

"As the rationals are dense in $\mathbb{R}$ we must have $\mathbb{R}\subset \bigcup_{n=1}^\infty I_n$." This is false. Pick your favorite irrational number $x$. For every $n$, there exist infinitely many rationals $q_m$ such that $q_m$ is within $a_n/2$ of $x$. But the least such $m$ may be much larger than $n$. In particular, it does not follow that there ...


32

Yes. Here is a proof over $\mathbb{C} $. Matrices with repeated eigenvalues are cut out as the zero locus of the discriminant of the characteristic polynomial, thus are algebraic sets. Some matrices have unique eigenvalues, so this algebraic set is proper. Proper closed algebraic sets have measure $0.$ (intuitively, a proper closed algebraic set is a ...


26

Another example: $$A=\mathbb{Z},\quad B=[0,1],\quad A+B=\mathbb{R}. $$


26

There are a couple of different issues going on here: if we increase the number of dimension ... the volume increases Here's the first problem: the volume of an $n$-dimensional cube whose side lengths are all $1$ is $1$ - for every $n$. These $1$s are not all the same thing, though, because they're measured in different units: e.g. if the side length is $...


25

Practically speaking this means that when you integrate functions such sets don't matter, in the sense that if you modify the domain of integration by a set of measure zero, the integral is unchanged.


20

Observe that the diameter of singletons is $0$ and the diameter of set $\{x,y\}$ is $d(x,y)>0$ if $x\neq y$. So there is no additivity.


18

Here is a slight variation on your construction. Consider $\mathbb{Q}$ which is countable, we may enumerate $\mathbb{Q}=\{q_1, q_2, \dots\}$. For each rational number $q_k$, cover it by an open interval $I_k$ centered at $q_k$ which does not contain $\pi.$ Each real number is arbitrarily close to a rational number since $\mathbb{Q}$ is dense in $\mathbb{R}$...


18

Another twist of the answer is: uniform subsets $A$ are almost nothing or almost everything. $m(A\cap(a,b))=(b-a)\,m(A)=m(A)\,m((a,b))$ implies $m(A\cap B)=m(A)\,m(B)$ for every measurable set $B$, due the the same $\sigma$-additivity and regularity of the measure. Now with $B=A$, we obtain $m(A)=m(A)^2$, meaning $m(A)=0$ or $m(A)=1$.


17

Well, if $f_k$ could be negative, then its integral might not even be defined. For instance, if $X=\mathbb{R}$ with Lebesgue measure and $f_k(x)=x$ for some $k$, there is no good way to define $\int f_k$ (it should morally be "$\infty-\infty$"). On the other hand, the integral of a nonnegative measurable function can always be defined (though it might be $\...


17

I preassume that here $A+B:=\{a+b\mid a\in A, b\in B\}$. Counterexample (for Lebesgue measure): Take $A=[0,1]\cup[2,3]$ and $B=[0,1]$ (so that $A+B=[0,4]$)


17

By Fubini, \begin{align*} I&=\int_{{\bf{R}}^{n}}e^{-a_{1}^{2}x_{1}^{2}}\cdots e^{-a_{n}^{2}x_{n}^{2}}dx_{1}\cdots dx_{n}\\ &=\int_{{\bf{R}}}\cdots\int_{\bf{R}}e^{-a_{1}^{2}x_{1}^{2}}dx_{1}\cdots e^{-a_{n}^{2}x_{n}^{2}}dx_{n}\\ &=\left(\int_{\bf{R}}e^{-a_{1}^{2}x_{1}^{2}}dx_{1}\right)\cdots\left(\int_{{\bf{R}}}e^{-a_{n}^{2}x_{n}^{2}}dx_{n}\right)\\...


15

Here is a nice proof due to Sierpiński, Démonstration élémentaire du théorème sur la densité des ensembles, Fundamenta Mathematicae, 4 (1), (1923), 167-171. I learned it from Appendix D in van Rooij, and Schikhof, A second course on real functions. The argument is indeed elementary, and applies to all sets, even non-measurable ones: Theorem. Given $E\...


15

Measures are defined on $\sigma$-algebras (containing the measurable sets) and any $\sigma$-algebra contains the empty set by definition.


15

It's not even finitely additive. If $X$ and $Y$ are two disjoint closed intervals on the real line then the diameter of their union is not the sum of their diameters.


15

If $\sum_{I \in X} l(I)=\infty$, then there is a countable subset of $X$ that sum up to infinity too. If $\sum_{I \in X} l(I)<\infty$, then only countably many terms of $l(I)$ is non-zero. In either case, it'd just be the same as allowing only countable infinite terms.


15

Hint: Show that the mapping $$(x,y) \mapsto T(x,y) := f(x)-y$$ is measurable. Conclude that $\Gamma(f) = T^{-1}(\{0\})$ is measurable.


14

Instead of trying to understand your approach (which sounds complicated), let me tell you how I'd do it. I guess you know how to construct a closed nowhere dense set of positive measure inside a given interval, right? Enumerate all the rational intervals in a sequence $I_1,I_2,I_3,\dots$. Now construct an infinite sequence $M_1,N_1,M_2,N_2,M_3,N_3,\dots$ of ...


14

Yes. Suppose that ${\frak m}^\ast E = 0$. On one hand, the inequality: $${\frak m}^\ast T \leq {\frak m}^\ast(T \cap E) + {\frak m}^\ast(T \cap E^c)$$always holds. On the other hand: $${\frak m}^\ast(T \cap E) \leq {\frak m}^\ast E = 0 \implies {\frak m}^\ast(T \cap E) = 0, \quad {\frak m}^\ast(T \cap E^c) \leq {\frak m}^\ast T$$gives the other inequality.


14

This part Each real number is arbitrarily close to a rational number since $\Bbb Q$ is dense in $\Bbb R$. Thus, each real number is in one of the open intervals. is not true. For an arbitrary $x\in\Bbb R$, $x$ doesn't necessary lies in any of $I_k$'s. If you are not convinced, you can try showing that there exists an $I_k$ such that $x\in I_k$, it wouldn'...


14

Consider the "melting ice cube" example (I like to have little mnemonics for remembering useful counterexmples): let $f_n := \frac{1}{n} \chi_{[0,n]}$. Then $f_n$ converges uniformly to zero, but $\int f_n = 1$ for all $n$, and so $$ \lim_{n\to \infty} \int f_n = 1 \ne 0 = \int \lim_{n\to\infty} f_n. $$ Note that if $\mu(X)$ is finite, then uniform ...


14

One can explicitly demonstrate the failure of the argument for $\mathbb R_+$. Generalize to $\mathbb R$ as needed. Consider the following maps $f:\mathbb N_+^2 \rightarrow \mathbb N_+$ and $g: \mathbb N_+^2 \rightarrow \mathbb Q_+$ \begin{eqnarray} f(\{k,m\}) &=& (k + m - 2 )(k + m - 1) + m \\ g(\{k,m\}) &=& k/m. \end{eqnarray} You can ...


13

Rather than removing things from the real line recursively like you do for Cantor sets, you could remove them all at once: Enumerate the rational numbers (or any other countable dense set) in [0,1] as $q_1, q_2, ...$, and let $B_n$ be an open interval of radius $\frac{1}{2^{n+1}}$ centered on $q_n$. Then $[0,1] - \bigcup B_n$ is a closed set (since it's ...


13

Since $f$ is strictly positive on $E$, we have $$ E = \bigcup_{n \geq 1} E_n, \quad \mbox{ where } E_n = \left\{x \in E: f(x) > \frac{1}{n}\right\}. $$ Since $\lambda(E) > 0$ there is some $n$ for which $\lambda(E_n)$ is positive (otherwise $E$ would be the countable union of measure $0$ sets, implying $\lambda(E)=0$). We then have $$ \int_E f \, d\...


13

A standard argument to show that $\mathcal{L}(\mathbb R)$ has the same size as $\mathcal{P}(\mathbb R)$ is to note that the Cantor subset of $[0,1]$ has the same size as $\mathbb R$ and measure 0, so any of its subsets also has measure 0. You can use the same idea to find the size of $\mathcal P(\mathbb R)\setminus\mathcal L(\mathbb R)$: Fix a nonmeasurable ...


13

The function is measurable because it is continuous and, also, positive. By the Fubini-Tonelli Theorem, $$ \begin{aligned}\int_{\mathbb{R}^n}\exp\left({-\sum_{i=1}^na_i^2x_i^2}\right)\,dx&=\prod_{i=1}^n\int_{\mathbb{R}}e^{-a_i^2x_i^2}\,dx_i=\prod_{i=1}^n\left(\int_{\mathbb{R}}e^{-a_i^2x^2}\,dx\right)\\&=\left(\frac{1}{|a_1\cdots a_n|}\int_{\mathbb{R}}...


12

This presentation is essentially repeated from "Measure and Integral" by Wheeden and Zygmund. The function $\lvert \cdot \rvert_e$ denotes outer measure. It can be proven that if $E$ is a measurable subset of $\mathbb{R}$ with positive measure, then the set of differences $\{ x-y \ \lvert \ x,y \in E \}$ contains an interval at the origin. The proof is in ...


12

Consider the set $B=e^{C} =\{e^x :x\in C\}.$ Then you have $$B\cdot B =e^C \cdot e^C =e^{C+C} .$$


12

Suppose there is such an $E.$ Clearly $m(E) > 0.$ Let $\epsilon>0.$ Then there are pairwise disjoint open intervals $I_1,I_2, \dots$ such that $E \subset \cup_n I_n$ and $\sum_n m(I_n) < m(E) +\epsilon.$ Thus $$m(E) = m(E\cap (\cup_n I_n)) = \sum_n m(E\cap I_n) = \sum_n \frac{m(I_n)}{2} < \frac{m(E)+\epsilon}{2}.$$ This implies $m(E) <\...


12

Under CH, there is such a set. To construct it, list all lines $\{L_i : i < \omega_1\}$ and construct $S$ inductively as follows. Start with empty $S$. At stage $i$, suppose $S$ is contained in $\bigcup \{L_j : j < i\}$ and it meets every $L_j$ at length one, for $j < i$. Note that $L_i \cap \bigcup \{L_j : j < i\}$ is countable. So we can choose ...


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